Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 6 - Inverse Functions - Review - Exercises - Page 505: 56


$$y' =1+\frac{1 }{y^2}$$

Work Step by Step

Given $$y=x+\tan^{-1}y$$ Differentiate both sides \begin{align*} y'&=1+\frac{y'}{1+y^2}\\ y'\left(1-\frac{1}{1+y^2}\right)&=1\\ y'\left( \frac{ y^2}{1+y^2}\right)&=1\\ y'&=\frac{1+y^2}{y^2}\\ &=1+\frac{1 }{y^2} \end{align*}
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